Verified Solution • 2025–26 Session moderate

QJEEM25Ma3069: JEE Main 2025 Mathematics - Class 12 JEEM25Ma3069 Solution: Let S = . Then is equal to _______.

Let S = {x:cos⁡−1x=π+sin⁡−1x+sin⁡−1[2x+1]}\left\{ x : \cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1} [2x + 1] \right\} . Then ∑x∈S(2x−1)2\sum\limits_{x \in S} (2x - 1)^2 is equal to _______.

Step-by-Step Verified Solution

amp;cos⁡−1x=π+sin⁡−1x+sin⁡−1(2x+1)amp;2cos⁡−1x−sin⁡−1(2x+1)=3π2amp;2α−β=3π2 where cos⁡−1x=α,sin⁡−1(2x+1)=βamp;2α=3π2+βamp;cos⁡2α=sin⁡βamp;2cos⁡2α−1=sin⁡β2x2−1=2x+1x2−x−1=0⇒x=1−52,{x=1+52 rejected }∴4x2−4x=4(2x−1)2=5\begin{aligned}& \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1) \\& 2 \cos ^{-1} x-\sin ^{-1}(2 x+1)=\frac{3 \pi}{2} \\& 2 \alpha-\beta=\frac{3 \pi}{2} \text { where } \cos ^{-1} x=\alpha, \sin ^{-1}(2 x+1)=\beta \\& 2 \alpha=\frac{3 \pi}{2}+\beta \\& \cos 2 \alpha=\sin \beta \\& \begin{array}{l}2 \cos ^2 \alpha-1=\sin \beta \\2 x^2-1=2 x+1 \\x^2-x-1=0 \\\Rightarrow x=\frac{1-\sqrt{5}}{2},\left\{x=\frac{1+\sqrt{5}}{2} \text { rejected }\right\} \\\therefore 4 x^2-4 x=4 \\(2 x-1)^2=5\end{array}\end{aligned}

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